MATHCOUNTS STATE COMPETITION SPRINT ROUND
2011MATHCOUNTS STATE COMPETITIONSPRINT ROUND1. 12 boy scouts are accompanied by 3scout leaders. Each person needs 3bottles of water per day and the trip is 1day.12 3 15 people15 3 45 bottles Ans.2. Cammie has pennies, nickels, dimesand quarters and we are asked to findthe least number of coins that she canuse to make 93 cents.Let’s start with quarters. 3 quartersequals 75 . 93 – 75 18 .We can use 1 dime. 18 – 10 8We can use 1 nickel. 8 – 5 3And we can use 3 pennies.3 1 1 3 8 Ans.3. We have 3 non-overlapping circles.One circle has a diameter of 8 inches.The other two circles have diameters of6 and 2 inches, respectively. We areasked to find how much larger the areaof the first circle is than the total area ofthe other 2 circles.If the diameter of the first circle is 8,then its radius is 4. Its area is 16S .The radii of the other two circles are 3and 1, respectively. Their areas are 9Sand S , respectively.16S 9S S6S Ans.4. Mike had 7 hits in 20 turns at bat in 5baseball games. In the sixth game hehad 5 hits in 5 turns at bat. We areasked to find the percent of Mike’s turnsat bat that resulted in hits.The total number of turns at bat is20 5 25 turnsThe total number of hits is7 5 12 hits122548% Ans.5. The first term of an arithmetic sequenceis –37 and the second is –30. We areasked to find the smallest positive termof the sequence.There is a difference of 7 between thetwo terms. If we look at the next 4 terms(–23, –16, –9, –2), we are adding 28 tothe second term to get –2 as the sixthterm.–2 7 5 Ans.6. Fraser fir tree saplings are 8 inches tall.Blue spruce tree saplings are 5 inchestall. Fraser firs grow at a constant rateof 12 inches per year and spruce treesat a constant rate of 14 inches per year.We are asked to find after how manyyears these trees will be the sameheight.Let n the number of years. Then,8 12n 5 14n3 2nn 3Ans.27. From the pie chart,1of my 303allowance was spent on movies,music,3on101on ice cream and the rest on5burgers.1 3 1 3 10 510 9 6302530or 25 out of 30 were spent on itemsother than burgers.30 – 25 5 Ans.8. M and N are both perfect squares lessthan 100. If M – N 27, then what isthe value of M N ?Let’s list all the squares less than 100.1, 4, 9, 16, 25, 36, 49, 64, 81, 100.If we add 27 to each square we get28, 31, 36, 43, 52, 63, 76, 91, 108 and127.36 is the only square. Therefore,M 9 and N 36M N9 36 3 6 9 Ans.9. John made two 120-mile trips. Thesecond trip took him one hour less thanthe first trip. Total time for both tripswas 9 hours. We are asked to find hisaverage rate for the second trip.If the second trip took him one hour less
than the first and the total number ofhours is 9, then the first trip was 5 hourslong and the second was 4 hours long.120430 Ans.10. We are asked to find the number ofparallelograms in the diagram.10 12 22 so far.We can also make parallelograms from3 small parallelograms. There are 3“horizontal” ones and 3 “vertical” ones.First the diagram itself contains 1parallelogram. Each of the smallestitems outlined in the parallelogram isitself a parallelogram.There are 9 of those for a total of 10 sofar.There are also 6 “horizontal”parallelograms that contain 2 smallparallelograms.And there are 6 “vertical” parallelogramsthat contains 2 small parallelograms.22 6 28We can also make parallelograms from4 parallelograms (2 adjacent and 2directly below). There are 4 of these.
28 4 32We can also make 2 “horizontal”parallelograms using 6 smallparallelograms and 2 “vertical” ones aswell.And that’s it. 32 4 36 Ans.11. 5 cm more than 3 times the length of arectangle is less than or equal to 44 cm.It’s also greater than or equal to 20 cm.The width of the same rectangle is 10cm. We are asked to find the positivedifference between the maximumpossible area of the rectangle and theminimum possible area of the rectangle.Let l the length of the rectangle.Let w the width of the rectangle.3l 5 443l 5 20w 10Let’s try and find the largest value of l.3l 5 443l 39l 13Therefore, 13 is the largest possiblevalue for the length and the area is13 10 130Now let’s try and find the smallest valueof l.3l 5 203l 15l 55 is the smallest possible length of therectangle and the smallest area is5 10 50130 – 50 80 Ans.12. Javier needs to exchange a dollar bill forcoins. The cashier has 2 quarters, 10dimes and 10 nickels. We are asked todetermine how many possiblecombinations of coins Javier could getback if he must get back at least one ofthe quarters.Let’s start by assuming he uses only 1quarter.D N716355That’s 5 combinations.4739Now, assume he’s going to get bothquarters.D N504234That’s 6 combinations.26180 10.5 6 11 Ans.13. There are four test scores which have amedian of 80 and a range of 12. We areasked to find the maximum possiblescore that could have been received ona test.Since there is an even number of tests,either both middle tests were 80 or theirsum was 2 80 160. The largestnumber we can choose for test #1 is 80so that the range will the maximumscore for test #4. Test #2 and #3 wouldthen be 80 each. Since the range is 12,then 80 12 92 is the largest testscore. 92 Ans.14. RS,TUV is a five-digit integer. It isdivisible by 5, S R2, and 10 T U 5. R, S, T, U and V are not necessarilydistinct. We must find how many fivedigit integers satisfy these conditions.If the five-digit integer is divisible by 5,then V 0 or V 5.
10 T U 5 must mean that T 0.and U 5. Otherwise the expressionwould be a value greater than 5.If S R2, then S could be 1 and R couldbe 1; or S could be 2 and R could be 4;or S could be 3 and R could be 9.So what does this all tell us?We can have:11050, 11055,42050, 42055,93050, 93055 or 6 integers. 6 Ans.215. y x 10x 21What is the least possible value of y?First, let’s find when y 0.x2 10x 21 0(x 3)(x 7) 0x –3, and x –7What happens when x –4?(–4)2 (10 –4) 21 16 – 40 21 –3What happens when x –5?(–5)2 (10 –5) 21 25 – 50 21 –4And what happens when x –6?(–6)2 (10 –6) 21 36 – 60 21 –3Since x –3 and x –7 give usexpressions that evaluate to 0, thesmallest value of the expression is–4. Ans.16. A square and a circle intersect so thateach side of the square contains achord of the circle which equals thecircle’s radius. We are asked to find theratio of the area of the square to thearea of the circle.2The area of the circle is Sr . If we drawa line from the center of the circle to thecircumference at the points where thechord is created, we have created anequilateral triangle of length r.2§r·h2 2¹3 2h2r43hr22h3rh2 r24r2223r3rThe area of the square isSo the ratio of the area of the square tothe area of the rectangle is3r 2Sr 23SAns.17. In rectangle ABCD, point E is on sideCD. The area of triangle ADE is1of5the area of quadrilateral ABCE. We areasked, what is the ratio of the length ofsegment DE to the length of segmentDC?Let l AD and let w DC.Then lw is the area of the rectangle.Let x DE.Let y the area of quadrilateral ABCE.Then6y51y y5lw.5lw6lw and yThe area of triangle ADE isxl21 5u lw5 61lw6xl23xl lw and 3x wTherefore, x The height of that triangle islength of the square.1the21Ans.31w and w is DC.318. The sum of the first n numbers in asequence of numbers is given by
n3 4n for all positive integers n. Weare asked to find the tenth number in thesequence.The first number is 13 4 5. That isactually the first number.The sum of the first 2 numbers is 8 8 16. The second number is 11.The sum of the first 3 numbers is 27 12 39. The third number is 23.The sum of the first numbers is 64 16 80. The fourth number is 51.So we have 5, 11, 23, 41. Can we seea pattern? I think so.11 – 5 623 – 11 1241 – 23 18Therefore, the fifth number is 41 24 65. The sixth is 65 30 95. Seventhis 95 36 131. Eighth is 131 42 173. Ninth is 173 48 221 and thetenth number is 221 54 275. Ans.19. We are given a square in the interior ofa hexagon and asked to find the degreemeasure of ABC.Hexagons have interior angles of 120º.Let s the side of the hexagon (and ofthe square).Then CD s and DB s making CDBan isosceles triangle. Since D 120ºand an angle of the square is 90º, BDC must be 30º. DCB DBC180 – 30 150Both DCB and DBC must then be75º. Since B is also 120º, ABC 120 – 75 45. Ans.20. When the diameter of a pizza increasedby 2 in, the area of the pizza increasesby 44%. We are asked to find the areaof the original pizza.Let x the radius of the original pizza.Then the area of the original pizza isSx 2 .Increasing the diameter by 2 in isthe same increasing the radius by 1 in.S x 1 2Sx 21.442x 11.44x2x 11.2xx 1 1.2 x0.2x 12x 10x 5The area of the original pizza is 25 S .Ans.21. When the numerator of a fraction isincreased by six, the value of thefraction increases by 1. If thedenominator of the fraction is increasedby 36, the value of the fractiondecreases by one.xbe the fraction. Thenyx 6 x 1 andyyxx 1y 36 yx 6 x yyyLetx 6 x yy 6Substituting into the second equation:xx 16 36 6xx 64266x 42(x – 6)x 7(x – 6)x 7x – 426x 42x 7The original fraction was7. Ans.622. The 80th term of an arithmetic sequenceis twice the 30th term. If the first term ofthe sequence is 7, then what is the 40thterm?The second term is 7 x, the third is 7 2x and so on. That makes the 30th term7 29x. The 80th term is 7 79x7 79x 2 (7 29x)
7 79x 14 58x121x 7 and x 3volume of the cone with radiusThe 40th term is 7 39x.7 39 x1·§7 39 u 3¹ 7 13 20 Ans.23. A right circular cone is sliced into fourpieces which have the same height. Weare asked to find the ratio of the volumeof the second largest piece to thevolume of the largest piece.Let r the radius of the cone andh the height of the cone.The radius grows proportionally as doesthe height so that for the smallest conethe radius is11r and the height is h ,44and for the cone made up of the first two1r and the height2pieces the radius isis1h , and so on for the other two2cones.The second largest piece is the volumeof the second largest cone (with radius33r and height h ) minus the volume44of the third largest cone (with radius11r and height h ).2221§3 · §3 ·V1uS u r h 3 4 ¹ 4 ¹19 39 2u S u u r 2hSr h316 46421§1 · §1 ·V2uS u r h 3 2 ¹ 2 ¹11 11 2u S u u r 2hSr h34 2249 21V1 V2Sr h Sr 2 h642427 2819 2Sr h Sr 2 hSr h192192192The volume of the largest piece is thevolume of the entire cone minus theheight3h , or V1 .43r and41 2Sr h31 29V3 V1Sr h Sr 2 h36464 227 237 2Sr h Sr hSr h192192192V3Taking the ratio:19 2Sr h19237 2Sr h192191923719219Ans.3724. x y z 7x2 y2 z2 19We are asked to find the mean of thethree products, xy, yz and xz.(x y z)2 72 49(x y z)2 x2 xy xz yx y2 yz zx zy z2x2 y2 z2 2xy 2xz 2yz 19 2(xy xz yz) 492(xy xz yz) 49 – 19 30xy xz yz 15xy xz yz31535 Ans.25. Quadrilateral ABCD is inscribed in acircle with segment AC a diameter ofthe circle. DAC 30º and BAC 45º.The ratio of the area of ABCD to the areaof the circle can be expressed as aa b. Wecʌare asked to find the value of a b c.common faction in the formBecause ADC and ABC eachintercept a semi-circle, each one is a90 angle. Therefore, ьACD is a30-60-90 triangle and ьABC is a
45-45-90 triangle. The hypotenuse ofeach triangle is 2r, where r is the radiusof the circle. The two legs of ьACD arer and r 3 (using 30-60-90 properties)3 21r . Ther(r 3 ) 22legs of ьABC are each r 2 (using45-45-90 properties) and its area is112(r 2 )2 (2r 2 ) r .22Therefore, the area of quadrilateral2 3 23 22ABCD isr r r . The22area of the circle is ʌ(r2). So, the ratio ofthe area of the quadrilateral to the areaof the circle is2 32 3 2r2 322 .and its area isʌʌU 22ʌThis makes a 2, b 3, and c 2.2 3 2 7 Ans.26. The numbers a, b, c and d form ageometric sequence.b a 3 and c b 9. We are askedto find the value of d.If c b 9, then c (a 3) 9 a 12.So we actually have the sequence:a, a 3, a 12, dIn a geometric sequence, each term is ntimes the previous term.a 3 n aa 3aa 3u a 3a 12aa 2 6a 9a 12ann9632a 3a3 323292323 24 27 2 22d n u a 12127 813u40 Ans.222a 1227. If you flip an unfair coin three times, theprobability of getting 3 heads is thesame as the probability of gettingexactly two tails. We are asked to findthe ratio of the probability of flipping atail to the probability of flipping a head.Let h the probability of a head comingup.Let t the probability of a tail coming up.We have one possibility that gives us 3heads, i.e., HHH and 3 possibilities ofexactly two tails, TTH, THT, and HTT.h3h2t2t3 u ht 23t 2h23h23h3h 33The ratio of the probability of flipping atail to the probability of flipping a head isthh 33h3Ans.328. The region shown is bounded by thearcs of circles having radius 4 units.Each arc has a central angle measure of60º and intersects at points of tangency.The area of the region can beexpressed in the form a b cS . Weare asked to find the value of a b c.a2 12a a2 6a 912a 6a 96a 9amultiply it by n.933Finally, now that we know what n is, let’sdetermine the value of a 12 andThis turns out to be pretty simple.If you draw lines connecting the three
centers of the circles, you end up withan equilateral triangle (that’s where the60º comes in) whose sides are twice theradius or 8. We must first find the areaof the triangle and for that we must havethe length of the height of the triangle.h 2 42 82h2 16 64h2 64 – 16 4848h4 3Therefore, the area of the triangle is:1u8u 4 3216 3From this, we need to subtract the areaof the sections of the circle formed bythe 60º central angles. But we have 3 ofthose which is equivalent to the areaformed by a semicircle. The area of the2circle is: Sr16S . Half of that is 8S .So, the area of the region bounded bythe arcs is 16 3 8S . This means:a 16, b 3, c –8a b c 16 3 – 8 11 Ans.29. A bag contains red and white balls. If 5balls will be pulled from the bag withreplacement, the probability of gettingexactly 3 red balls is 32 times theprobably of getting exactly one red ball.We are asked to find what percent of theballs originally in the bag are red.Let r the number of red balls.Let w the number of white balls.Then the probability of picking a red ballrand ther wfrom the bag isprobability of picking a white ball fromthe bag isw.r wThe probability of getting 3 red balls is:r 3 w2r w5u 10r wr 3 w 2 16rw 4r 2 16w 2r 4w32rw 45The probability of getting only one redrw 4r w5!There are4!1!55 ways to do it.5r wu5So, there are 4 more times as many redballs as white balls. That’s 80%. Ans.30. 52,683 52,683 – 52,660 52,706 (52660 23)2–(52660 (52660 46)) 526602 232 (2 23 52660) –(526602 (52660 46)) 232 (46 52660) – (46 52660) 232 529 Ans.TARGET ROUND1. We are asked to find the smallestpossible integer such that none of thedigits is a zero, the integer is a multipleof 3 and exactly one of the digits is aperfect square.A multiple of 3 means that if you add thedigits in the number, the sum is itself amultiple of 3.The first perfect square is 1. We musthave 2 other digits that are not perfectsquares. That would have to be 2 and 3so the smallest number is 123. Ans.2. We have a truck that is driven 480 milesat 60 miles per hour. The truckaverages 20 miles per gallon. Thedriver is paid 12/hour. Gasoline costs 3 per gallon Finally, maintenance onthe truck is 0.15 per mile. We mustdetermine what the total cost of the tripis for the company.480 miles at 60 miles per hour meansthat the truck is driven for5!There are10 ways to do it.3!2!ball is:r 3w2480608hours. So the labor cost is12 8 96. 480 miles at 20 miles pergallon means we must buy48020gallons of gas at 3 per gallon.24 3 72.96 72 168, so far.Is there anything else? Yes, themaintenance.24
480 0.15 72168 72 240 Ans.11of the pizza. Marti had541ofof what was left. Then Bobbi had33. Jose hadthe remaining pizza and after that, Ms.Gauss took1of that. So how much2pizza is left?1of the pizza there was51 41of the pizza left. Marti ate1 5 541 4 1of the pizza. Nowof that or u4 5 54 1 3of the pizza is left. Bobbi 5 5 511 3 1of that or uof thegets33 5 53 1 2of thepizza which leaves 5 5 5After Jose hadpizza left. Finally Ms. Gauss gets her11 2 1of the leftovers or uof the22 5 52 1 1of thepizza. That leaves 5 5 51or 20% Ans.pizza left.54. Marco will celebrate his Nth birthday inthe year N2, which is in the 21st century.So, in what year did he celebrate his13th birthday?Okay, we have to find a square that isbetween 2000 and 2100.402 1600 and 502 2500 We’relooking for a value between 40 and 50,probably somewhere in the mid 40’srange. 452 2025 and that must be it!So, Marco will be 45 in 2025.45 – 13 32 which means 32 yearsbefore 2025, Marco was 13.2025 – 32 1993 Ans.5. How many distinct sums can beobtained by adding three differentnumbers from the set {3, 6, 9, , 27,30}?Let’s rewrite that as{3 1, 3 2, 3 3, , 3 9, 3 10}And we can reduce this to how manyvalues can you get adding three of thenumbers 1 through 10.There are 10 9 8 720 differentvalues. But many of them are the same.The smallest value is 1 2 3 6.The largest value is 8 9 10 27.And we can get every value in between.27 – 6 21 But don’t forget to count the6, because there are actually 22numbers from 6 to 27. 22 Ans.6. x and y are real numbers such thatxy 9 and x2y xy2 x y 100We are asked to find the value ofx2 y2x2y xy2 x y 100x xy y xy x y 1009x 9y x y 10010(x y) 100x y 10(x y)2 100 x2 2xy y2x2 y2 (2 9) 100x2 y2 100 – 18 82 Ans.7. In a shooting competition, the object ofthe match is to be the first to hit thebull’s eye of a target 100 feet away.Each opponent has a 40% chance ofhitting the bull’s-eye on a given shot.Franz shoots first and we mustdetermine the probability that Hans willwin the competition and take no morethan 3 shots.Scenario 1: Franz misses and thenHans hits the bull’s-eye.The probability of this is0.6 0.4 0.24Scenario 2: Franz misses, Hans misses,Franz misses and Hans hits the bull’seye.0.6 0,6 0.6 0.4 .0864Scenario 3: Franz misses, Hans misses,Franz misses, Hans misses, Franzmisses and Hans hits the bull’s-eye.0.6 0.6 0.6 0.6 0.6 0.4 0.031104The probability of all 3 scenarios is:0.24 0.0864 0.031104 0.357504 § Ans.
8. The solid figure has 6 faces that aresquares and 8 faces that are equilateraltriangles. Each of the 24 edges haslength 2 cm. The volume of the solidcan be expressed in the formac where c has no perfect squarebfactor except 1, and a and b arerelatively prime. We must find the valueof a b c.OK. I’m not going to even try toreproduce that figure so I’m just going totalk through it. First of all, there is arectangular solid in there whose lengthand width are each 2 cm and whoseheight is the diagonal of a square whoseside is 2 cm. The diagonal must be2 2 and the volume of the solid is2 u 2 u 2 2 8 2.Now, if you remove that part of theobject, you’re left with 4 halfoctahedrons. An octahedron looks likethis:We actually, then, have 2 fulloctahedrons. The volume of anoctahedron with side 2 is1u 2 u 233823The volume of two of them is8 2 16234023162.3This means that a 40, b 3 and c 2.a b c 40 3 2 45 Ans.TEAM ROUND1. Kevin, Cindi and Marcus have 1020widgets. Marcus has half of what Cindihas and Kevin has 219 widgets. Sohow many widgets does Cindi have?Let x the number of widgets thatMarcus has. Then Cindi has 2xwidgets.1020 – 219 801x 2x 8013x 801x 2672x 534 Ans.2. We have a three-digit integer where alldigits are prime and distinct. They alsoincrease in order from left to right. Weare asked to determine how
MATHCOUNTS STATE COMPETITION SPRINT ROUND 1. 12 boy scouts are accompanied by 3 scout leaders. Each person needs 3 bottles of water per day and the trip is 1 day. 12 3 15 people 15 3 45 bottles Ans. 2. Cammie has pennies, nickels, dimes and quarters and we are asked to find the
Table 2 showed the time for each sprint in RSA without ball conditions. Sprint 1 and Sprint 2 were shown to be significantly faster compared to Sprint 5 (Sprint 1, p 0.012; Sprint 2, p 0.038). Table 2. Sprint time without ball Sprint RSA time without ball (s) Mean SD Sprint 1 7.22 0.62 Sprint 2 7.22 0.64 Sprint 3 7.24 0.65
2018 AMC 8 Problem 2 exactly the same as 2012 3 is MathCounts State Sprint Problem 3, and very similar to the following 4 problems: 2017 MathCounts Chapter Countdown #49 2016 MathCounts National Sprint #11 2016 - 2017 MathCounts School
math team placed fifth out of 32 middle schools competing in the MATHCOUNTS 2006 Oahu Chapter Competition, and are two of the six Oahu math teams that will represent Oahu in the Hawaii State MATHCOUNTS Competition on March 11, 2006. Other public . The first is the Sprint Round with its 30 problems, and then comes the Target
ten Sprint and Target Rounds determine if he or she competes in the Countdown Round, the oral portion of the competition. The ten highest scorers advance to the Countdown Round with the top four winners representing the State of Hawaii at the National MathCounts Competition on May 7, in Washington, D.C. The top four winners in the Countdown
first place in the 2012 MATHCOUNTS 29th Hawaii State Competition on March 3, 2012 held at Iolani School. A week earlier . Round, followed by the Target, Team, and Countdown Rounds. The Sprint Round has individual Mathletes answering 30 difficult problems in 40 minutes. The Target Round has
2021 MATHCOUNTS TEXAS STATE COMPETITION 2 2021 March 25, 2021 - Virtual State MATHCOUNTS Competition The State Competition consists of the Sprint and Target Rounds and will take place online on the Art of Problem Solving Contest Platform at 6:00pm CST on Thursda
SPRINT 7 - 18 NEEdLES SPRINT 7XL - 18 NEEdLES SPRINT 7 L - 18 NEEdLES SPrINT 7 the SPrint 7 Series is equipped with 18 needles and servo drives. the sprint 7 is ca. 7,5% more efficient than the sprint 6 due to increased speeds at longer stitch lengths. the benchmark of 18 needles increases efficiency by reduction
unstoppable: a few mouthfuls packed enough nutritional punch to let them run all day without rest. But whatever secrets the Tarahumara are hiding, they’ve hidden them well. To this day, the Tarahumara live in the side of cliffs higher than a hawk’s nest in a land few have ever seen. The Barrancas are a lost world in the most remote wilderness in North America, a sort of a shorebound .
